Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int n, m, q, p[100002][20], hi[100002];
vector<int> v[100002], id[100002][20], ans;
void dfs(int node, int par, int h) {
  p[node][0] = par;
  hi[node] = h;
  for (int i = 0; i < v[node].size(); i++)
    if (v[node][i] != par) dfs(v[node][i], node, h + 1);
}
void mergef(vector<int>& a, vector<int>& b, vector<int>& c, int k) {
  int cntb = 0, cntc = 0;
  vector<int> a_dum;
  for (int i = 0; i < k; i++) {
    if (cntb < b.size() && cntc < c.size()) {
      if (b[cntb] < c[cntc])
        a_dum.push_back(b[cntb++]);
      else
        a_dum.push_back(c[cntc++]);
    } else if (cntb < b.size())
      a_dum.push_back(b[cntb++]);
    else if (cntc < c.size())
      a_dum.push_back(c[cntc++]);
  }
  a = a_dum;
}
void precompute_lca_and_top10() {
  for (int i = 1; (1 << i) <= n; i++)
    for (int node = 1; node <= n; node++) {
      p[node][i] = p[p[node][i - 1]][i - 1];
      mergef(id[node][i], id[node][i - 1], id[p[node][i - 1]][i - 1], 10);
    }
}
void top_k(int a, int b, int k) {
  int log;
  for (log = 1; (1 << log) <= hi[a]; log++)
    ;
  log--;
  for (int i = log; i >= 0; i--)
    if (hi[a] - (1 << i) >= hi[b]) {
      mergef(ans, ans, id[a][i], k);
      a = p[a][i];
    }
  if (a == b) {
    mergef(ans, ans, id[b][0], k);
    return;
  }
  for (int i = log; i >= 0; i--) {
    if (p[a][i] != p[b][i]) {
      mergef(ans, ans, id[a][i], k);
      mergef(ans, ans, id[b][i], k);
      a = p[a][i], b = p[b][i];
    }
  }
  mergef(ans, ans, id[a][1], k);
  mergef(ans, ans, id[b][0], k);
}
int main() {
  int a, b, k;
  scanf("%d%d%d", &n, &m, &q);
  for (int i = 1; i < n; i++) {
    scanf("%d%d", &a, &b);
    v[a].push_back(b);
    v[b].push_back(a);
  }
  for (int ids = 1; ids <= m; ids++) {
    int node;
    scanf("%d", &node);
    id[node][0].push_back(ids);
  }
  for (int i = 1; i <= n; i++) {
    sort(id[i][0].begin(), id[i][0].end());
    while (id[i][0].size() > 10) id[i][0].pop_back();
  }
  dfs(1, 0, 1);
  precompute_lca_and_top10();
  while (q--) {
    ans.clear();
    scanf("%d%d%d", &a, &b, &k);
    if (hi[a] < hi[b]) swap(a, b);
    top_k(a, b, k);
    printf("%d ", (int)ans.size());
    for (int i = 0; i < (int)ans.size(); i++) printf("%d ", ans[i]);
    printf("\n");
  }
  return 0;
}